Chapter 1.1, Problem 14E

Given lim x-4 f (x) = 1,limx-49 (x) = 10, and lim→-4 h (x) = -7 use the limit properties to find lim→-4 1 [2h (x) — h(x) + 7 f(x)] : - h(x)+7f(x) 3 O DNE

17. Suppose we know that the graph below is the graph of a solution to dy/dt = f(t). (a) How much of the slope field can you sketch from this information? [Hint: Note that the differential equation depends only on t.] (b) What can you say about the solu- tion with y(0) = 2? (For example, can you sketch the graph of this so- lution?) y(0) = 1 y AN

(b) Find the (instantaneous) rate of change of y at x = 5. In the previous part, we found the average rate of change for several intervals of decreasing size starting at x = 5. The instantaneous rate of change of fat x = 5 is the limit of the average rate of change over the interval [x, x + h] as h approaches 0. This is given by the derivative in the following limit. lim h→0 - f(x + h) − f(x) h The first step to find this limit is to compute f(x + h). Recall that this means replacing the input variable x with the expression x + h in the rule defining f. f(x + h) = (x + h)² - 5(x+ h) = 2xh+h2_ x² + 2xh + h² 5✔ - 5 )x - 5h Step 4 - The second step for finding the derivative of fat x is to find the difference f(x + h) − f(x). - f(x + h) f(x) = = (x² x² + 2xh + h² - ])- = 2x + h² - 5h ])x-5h) - (x² - 5x) = ]) (2x + h - 5) Macbook Pro